\((1) \Longleftrightarrow (2)\)  follows by combining Lemme 4.4 and Corollaire 4.15 in [1]}.
\((3) \Rightarrow (1)\) : if there exists a formal model \(\mathfrak {X}\)  as in (3), then taking \(W \subset X\)  be the preimage of any formal affine open \(\mathfrak {X}\)  containing the image of \(x\)  shows that \(x\)  is weakly Shilov.
\((1) \Rightarrow (3)\) : Assume \(x \in X\)  is weakly Shilov. We can then find an affinoid open \(\mathrm {Spa}(R) \subset X\)  containing \(x\)  such that \(x \in \mathrm {Spa}(R)\)  is Shilov. By Raynaud, there is a formal model \(\mathfrak {X}\)  of \(X\)  such that \(\mathrm {Spa}(R) \subset X\)  is the preimage of a formal affine open \(\mathrm {Spf}(R_0) \subset \mathfrak {X}\) , cf. [2]}. By passing to a refinement, we can assume that the map \(R_0/t \rightarrow \tilde{R}\)  gives a homeomorphism of spectra; see first paragraph of the proof of Proposition REF . As \(x\)  is weakly Shilov, its image in \(\mathrm {Spec}(\tilde{R})\)  is a generic point. But then its image in \(\mathrm {Spec}(R_0/t)\)  is also a generic point since \(\mathrm {Spec}(\tilde{R}) \rightarrow \mathrm {Spec}(R_0/t)\)  is a homeomorphism by construction. As \(\mathrm {Spf}(R_0) \subset \mathfrak {X}\)  is a formal open immersion, it follows that \(x\)  gives a generic point of \(\mathfrak {X}_s\)  as well
